document.write( "Question 127046: \r
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document.write( "Janet invested $31,000, part at 10% and part at 1%. If the total interest at the end of the year is $1,390, how much did she invest at 10%?\r
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document.write( "please help \n" );
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Algebra.Com's Answer #93062 by checkley71(8403)![]() ![]() ![]() You can put this solution on YOUR website! .10x+.01(31,000-x)=1390 \n" ); document.write( ".10x+310-.01x=1390 \n" ); document.write( ".09x=1390-310 \n" ); document.write( ".09x=1080 \n" ); document.write( "x=1080/.09 \n" ); document.write( "x=12,000 answer for the amount invested @ 10%. \n" ); document.write( "31,000-12,000=19,000 amount invested @ 1%. \n" ); document.write( "PROOF: \n" ); document.write( ".10*12000+.01*19000=1390 \n" ); document.write( "1200+190=1390 \n" ); document.write( "1390=1390 \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |